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PROBLEM SET 10
Solutions
Problem 1
- • The Bode gain is KBode = 1, and there are no integrators, so the low-frequency asymptote is
MdB = 0. Again because there are no integrators, the phase at low frequencies is 0 .
- The break frequency of the pole is � = 1. At this frequency, the slope of the magnitude plot
changes by -20 dB/dec. The phase changes by − 5 ^ one decade before the break frequency (at � = 0 .1), by − 45 ^ at the break frequency (at � = 1), and by − 85 ^ one decade after the break
frequency (at � = 10). At high frequencies, the phase approaches − 90 .
G 1
(s)
Phase (deg)
Magnitude (dB)
−
−
−
−
−
0
−
−
0
−1 0 1 10 10 10
Frequency (rad/sec)
- • KBode = 1 and there is one integrator, so the low-frequency asymptote has a slope of -20 dB/dec
and a magnitude of 1 (= 0 dB) at � = 1. Because there is one integrator, the phase at low
frequencies is − 90 .
- The pole at s = − 1 breaks at � = 1, so the slope of the magnitude plot changes by -20 dB/dec
at that point. (It goes from -20 dB/dec to -40 dB/dec.) The phase is approximately − 95 ^ at � = 0.1, − 135 ^ at � = 1, and − 175 ^ at � = 10. The phase at high frequencies approaches
− 180 ^.
G 2
(s)
Phase (deg)
Magnitude (dB)
−
−
−
−
−
−
0
10
20
−
−
−
−1 0 1 10 10 10
Frequency (rad/sec)
- • KBode = 10 and there are no integrators, so the low-frequency asymptote is MdB = 20 and the
phase at low frequencies is zero.
- The first break frequency is due to the pole at s = 1. So at � = 1 the slope of the magnitude
Phase (deg)
Magnitude (dB)
plot changes from 0 dB/dec to -20 dB/dec. Then the break frequency due to the zero is at
� = 10, so at that frequency the slope changes from -20 dB/dec back to 0 dB/dec.
- At � = 0 .1, the pole contributes − 5 ^ and the zero contributes nothing, so the phase is − 5 .
- At � = 1, the pole contributes − 45 ^ and the zero contributes +5, so the phase is − 40 .
- At � = 10, the pole contributes − 85 ^ and the zero contributes +45, so the phase is − 40 .
- At � = 100, the pole contributes − 90 ^ and the zero contributes +85, so the phase is − 5 .
- As � ⇒ ≈, the phase contribution from the pole and zero cancel and so the phase is 0 .
G 4
(s)
20
15
10
5
0
0
−
− −1 0 1 2 10 10 10 10
Frequency (rad/sec)
- • KBode = 0. 1 and there are no integrators, so the low-frequency asymptote is MdB = − 20 and
the phase at low frequencies is zero.
- The first break frequency is due to the zero at s = 1. So at � = 1 the slope of the magnitude plot changes from 0 dB/dec to +20 dB/dec. Then the break frequency due to the pole is at
� = 10, so at that frequency the slope changes from +20 dB/dec back to 0 dB/dec.
- At � = 0 .1, the zero contributes +5^ and the pole contributes nothing, so the phase is +5.
- At � = 1, the zero contributes 45 ^ and the pole contributes − 5 , so the phase is +40.
- At � = 10, the zero contributes 85 ^ and the pole contributes − 45 , so the phase is +40.
- At � = 100, the zero contributes 90 ^ and the pole contributes − 85 , so the phase is +5.
- As � ⇒ ≈, the phase contribution from the pole and zero cancel and so the phase is 0 .
G 5
(s)
Phase (deg)
Magnitude (dB)
−
−
−
−
0
0
30
60
−1 0 1 2 10 10 10 10
Frequency (rad/sec)
�
Problem 2
Bode Diagram Gm = 28.696 dB (at 4.0818 rad/sec), Pm = 83.577 deg (at 0.24932 rad/sec) 50
0
Phase (deg)
Magnitude (dB)
−
−
−
−
−
−
−
−
−
− 10 −2 10 −1 100 101 102 Frequency (rad/sec)
PM = <BOC
GM = 1 / |OB|
O
C
-1 B
w = 0
w = 0+
�
Problem 3
- The first thing to notice is that the low-frequency asymptote is MdB = 20 dB and the phase at low frequencies is zero, so there are no integrators and the Bode (standard) gain of G 1 (s) is 10. At
around � = 1, the slope of the magnitude plot drops to -20 dB/dec and the phase turns negative, so
there is a pole at s = −1. Then at around � = 30, the slope of the magnitude plot reverts to zero, and the phase starts getting more positive, so there is a zero at s = −30. Finally, at � = 1000, the
slope drops to -20 dB/dec again and the phase gets more negative, so there is a pole at s = −1000. The final transfer function is then:
10( 30 s + 1) G 1 (s) = (^) s (s + 1)( 1000 + 1)
Check: there is one more pole than zero, so the phase at very high frequencies should be − 90 ^ (it is).
- Now the low-frequency asymptote has a slope of -20 dB/dec and a magnitude of 0 dB at � = 1, and the phase at very low frequencies is − 90 . So there is one integrator, and the Bode gain is 1.
At � = 1, the slope changes to 0 dB/dec, indicating the presence of a zero. However, the phase changes by − 90 , so the zero must be in the right half-plane at s = 1.
Then at � = 100, the slope changes to -20 dB/dec, indicating the presence of a pole. However, the phase changes by +90^ so the pole must be in the right half-plane at s = 100.
� G 2 (s) =
−s + 1
s(− s 100 +^ 1)
Problem 4
- The crossover frequency is the frequency at which |G (j�) |= 1 or 0 dB. The gain margin is the
difference in magnitude between 0 dB and |G (j�) |when the phase is − 180 ^. The phase margin is the difference in phase between − 180 ^ and �G(j�) when the magnitude is 1. From the tabulated
data (or the Bode plot), we get:
�c � 3. 1 1 GM �
- 25
�m 180
− 135
= 45
From Figure 8.7: a phase margin of 45 ^ corresponds to a damping ratio of � � 0 .43. From Figure �c 8.6: this damping ratio corresponds to (^) �n � 0 .84, so �n =
1
84 =^3 .7.
From Table 8.4.1 (or Figure 8.7): to get � = 0 .6, we need �m = 59 .
To get this phase margin, we want the phase of G(j�) to be − 180 ^ + 59 ^ = − 121 ^ when |G (j�) =| 1.
From the tabulated data, G(j�) has this phase at � = 2. 46 and |G (j�) |= 1. 42 at this frequency. So we need to multiply G(j�) by a factor of 1 / 1. 42 =. 704 to get the desired phase margin. So
K = .704.
- From Table 8.4.1: � = 0. 6 � ��c^ = 0.72. The new crossover frequency is 2.46, so the new natural n frequency is �^2.^46 n =^0. 72 =^3 .4.
